A2 June 2019 Paper 2 Q4
4 A 2-D transformation T is a shear which leaves the \(y\)-axis invariant and which transforms the object point \((2, 1)\) to the image point \((2, 9)\). \(\mathbf{A}\) is the matrix which represents the transformation T.
(a) Find \(\mathbf{A}\). [3]
(b) By considering the determinant of \(\mathbf{A}\), explain why the area of a shape is invariant under T. [2]
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{A} = \begin{pmatrix} 1 & 0 \\ k & 1 \end{pmatrix}\) | B1 | 1.2 |
| \(\mathbf{A} = \begin{pmatrix} 1 & 0 \\ k & 1 \end{pmatrix}\begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 2k + 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 9 \end{pmatrix}\) | M1 | 1.1 |
| \(k = 4\) so \(\mathbf{A} = \begin{pmatrix} 1 & 0 \\ 4 & 1 \end{pmatrix}\) | A1 | 1.1 |
| [3] |
Notes
B1: Correct form of \(\mathbf{A}\) seen
M1: Correctly multiplying object vector into their \(\mathbf{A}\) to find image and equating to given image. Their \(\mathbf{A}\) must have at least 1 unknown element
| Scheme | Marks | AO |
|---|---|---|
| \(\det\mathbf{A} = 1 - 0 = 1\) | B1 | 1.1 |
| The determinant is the area scale factor so a determinant of 1 leaves the area unchanged | B1 | 2.4 |
| [2] |
Notes
B1: Correctly finding determinant
B1: Convincing explanation. Must include both ideas.